In the proof here a strictly positive function in $(0,\pi)$ is integrated over this interval and the integral is claimed as a positive number. It seems intuitively obvious as the area enclosed by a continuous function's graph lying entirely above the x-axis and the x-axis should not be zero. But how can I prove this formally?

If the function is positive over a closed interval apparently the result is not true (link goes to page 147 in Theories of Integration by Kurtz and Swarz). This has further confused me. Can someone please clarify my doubt.


  • 1
    $\begingroup$ If you have a counterexample function $f$ s.t. $f(x)>0$ on the closed interval $[a,b]$ but $\int_a^b f = 0$, then clearly $f(x)>0$ on $(a,b)$ and $\int_a^b f = 0$. $\endgroup$ – Ilya Apr 4 '13 at 14:53
  • $\begingroup$ Any graph argument implicitly assumes that the function is continuous. In which case $f>0$ implies $\int_a^b f>0$. $\endgroup$ – Julien Apr 4 '13 at 14:53
  • $\begingroup$ to add to a comment by @julien, although Riemann integrability does not require continuity, the function in your first link is continuous $\endgroup$ – Ilya Apr 4 '13 at 14:56
  • $\begingroup$ If $f$ is Riemann integrable, then $f$ is continuous a.e., and one just needs continuity at a single point to show $\int_a^b f > 0$. $\endgroup$ – copper.hat Apr 4 '13 at 16:10

Let $f : I \to \mathbb{R}$ be a function on some interval $I$.

If $f$ is continuous and positive on $I$, then $\displaystyle \int_I f >0$. Indeed, $f \geq \alpha >0$ on some closed interval $K \subset I$, so $\displaystyle \int_I f \geq \int_K f \geq \alpha \cdot \mu(K)>0$.

  • $\begingroup$ @julien But Riemann integrable $\implies$ Lebesgue integrable, hence how could a counterexample exist? $\endgroup$ – Did Apr 4 '13 at 15:13
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    $\begingroup$ A Riemann integrable function $f$ over a nondegenerate interval has points of continuity. This is enough to insure the integral is positive for positive $f$. $\endgroup$ – David Mitra Apr 4 '13 at 15:15
  • $\begingroup$ @DavidMitra I was impressed by Remark 4.21 here. And I did not really think about it any further... $\endgroup$ – Julien Apr 4 '13 at 15:19
  • $\begingroup$ @Did Sure, I was silly. Do you have any idea of what they mean here in Remark 4.21? $\endgroup$ – Julien Apr 4 '13 at 15:29
  • $\begingroup$ @julien: I can only imagine that they mean with the machinery developed so far, they cannot prove it. However, I think it is straightforward to prove, let me see if I can dig up a proof. $\endgroup$ – copper.hat Apr 4 '13 at 15:41

Here is a "direct" proof that does not require knowledge of the continuity properties of Riemann-integrable functions:

  • Suppose by way of contradiction that $\int_0^1 f = 0$.
  • It follows that the upper Darboux sums can be made arbitrarily small. Choose some partition $\mathcal{P}_1$ such that $U(f; \mathcal{P}_1) < 1$.
  • There must be some subinterval $I_1$ of $\mathcal{P}_1$ on which the supremum of $f$ is less than 1, as otherwise, summing over all subintervals contradicts the condition that $U(f; \mathcal{P}_1)< 1$.
  • Choose some partition $\mathcal{P}_2$ for which $U(f; \mathcal{P}_2) < \frac{|I_1|}{2}$. We may assume WLOG that $\mathcal{P}_2$ is a refinement of $\mathcal{P}_1$, as otherwise re replace $\mathcal{P}_2$ by their common refinement.
  • Of the subintervals of $I_1$ which are in $\mathcal{P}_2$, their must be one, which we call $I_2$, on which the supremum of $f$ is less than $\frac{1}{2}$. If not, the sum of $M_k \Delta x_k$ over the subintervals of $I_1$ would yield at least $\frac{|I_1|}{2}$, and $U(f; \mathcal{P}_2)$ would be at least this much.
  • Similarly, choose $\mathcal{P}_3$ a refinement of $\mathcal{P}_2$ with the property $U(f; \mathcal{P}_3) < \frac{|I_2|}{3}$.
  • Continuing in this manner, we obtain a nested sequence $I_1 \supseteq I_2 \supseteq \dots$ of closed bounded intervals with the property that $f(x) < \frac{1}{n}$ for all $x \in I_n$.
  • By the Nested Interval Property, the intersection $\bigcap_{n=1}^\infty I_n$ is nonempty. But any point $x$ in this intersection must have the property $\forall n \in \mathbb{N}: f(x) < \frac{1}{n}$, which implies $f(x) = 0$.

If $f$ is non-negative, Riemann integrable and $\int_a^b f(x) dx = 0$ (with $a<b$) then it must be the case that $f(x) = 0$ a.e. Hence if $f$ is strictly positive on $[a,b]$, then it must be the case that $\int_a^b f(x) dx > 0$.

This is straightforward to see using the Lebesgue integral.

See Corollary 3 in www.math.sc.edu/~schep/riemann.pdf for a straightforward proof. The essence is to show that for all $c > 0$ the set $\{x \in [a, b] | f (x) \ge c \}$ has content zero.


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